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🧪 NEET Chemistry❓ Question

Finding the rate law from a data table, unit of k, and order vs molecularity. I keep losing marks here

Late night again, sorry if the formatting is bad.

Mehak AroraDropper Sample

20 Sept 2026 · 1 min read

Chemical kinetics is the chapter where my anxiety goes up, because every question has some tiny trap. Let me show the problem I did and where my confusion is.

Data for a reaction between A and B (concentrations in mol/L, rate in mol L⁻¹ s⁻¹):

Exp[A][B]Rate
10.100.102.0×10−32.0\times10^{-3}
20.200.108.0×10−38.0\times10^{-3}
30.100.204.0×10−34.0\times10^{-3}

I compared 1 and 2: [A] doubled and rate became four times, so order in A is 2. Comparing 1 and 3: [B] doubled, rate doubled, order 1. Overall 3. Is that right? And then what is the unit of kk? I wrote s⁻¹ and was marked wrong.

Other doubts:

  1. For 2N2O5→4NO2+O22\text{N}_2\text{O}_5 \to 4\text{NO}_2 + \text{O}_2 the rate of reaction can be written using any species. How does the coefficient come in? I always put the wrong fraction.
  2. What's the difference between order and molecularity? Can the order be zero or a fraction?

Please tell me if my method is OK. I want to be sure before the weekend test.

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6 answers & replies

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  • [+]Zoya Khan12thSample· 29d ago

    Your method is exactly right, and your answer too: rate =k[A]2[B]= k[A]^2[B], overall order 3. Find kk from experiment 1:

    k=2.0×10−3(0.10)2(0.10)=2.0 L2mol−2s−1k = \frac{2.0\times10^{-3}}{(0.10)^2(0.10)} = 2.0\ \text{L}^2\text{mol}^{-2}\text{s}^{-1}

    You can check with experiment 2: 2.0×(0.2)2×0.1=8.0×10−32.0 \times (0.2)^2 \times 0.1 = 8.0\times10^{-3} ✓.

    For (1): divide each concentration change by its stoichiometric coefficient so that all give the same number:

    rate=−12d[N2O5]dt=14d[NO2]dt=d[O2]dt\text{rate} = -\frac12\frac{d[\text{N}_2\text{O}_5]}{dt} = \frac14\frac{d[\text{NO}_2]}{dt} = \frac{d[\text{O}_2]}{dt}

    NO₂ appears four times as fast as O₂ does, so you divide by 4.

  • [+]Lakshmi TripathiDropperSample· 29d ago

    Unit of kk: derive it, don't memorise. k=rate[A]2[B]k = \dfrac{\text{rate}}{[A]^2[B]}, so units are mol L−1s−1(mol L−1)3=L2 mol−2 s−1\dfrac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^3} = \text{L}^2\,\text{mol}^{-2}\,\text{s}^{-1}. General: for order nn, the unit is (mol L−1)1−ns−1(\text{mol L}^{-1})^{1-n}\text{s}^{-1}. So zero order: mol L⁻¹ s⁻¹, first order: s⁻¹, second: L mol⁻¹ s⁻¹, third: L² mol⁻² s⁻¹. 'Only first order has plain s⁻¹'.

    • [+]Mehak AroraDropperSample· 29d agoAuthor

      Deriving the unit instead of memorising is such a calming idea. Thank you all, I feel like I can actually sleep now. 💛

  • [+]Ishaan Bose11thSample· 29d ago

    Order is just the sum of the powers in the balanced equation. For 2N₂O₅ → 4NO₂ + O₂ the order is 2 because of the 2 in front, and molecularity can be anything including fractions, like 1.5, depending on the mechanism. And order must be a whole number, that's the rule. Fractional order is a myth by chemists who want to confuse us 😅

    • [+]Siddharth RaoDropperSample· 29d ago

      Ishaan, a few things there are the other way round. Order comes from the experiment, not the stoichiometry. Decomposition of N₂O₅ is actually first order, even with the 2 in front, as the rate depends on [N2O5]1[\text{N}_2\text{O}_5]^1. Order can be zero or a fraction: NCERT gives CH3CHO→CH4+CO\text{CH}_3\text{CHO} \to \text{CH}_4 + \text{CO} with rate =k[CH3CHO]3/2= k[\text{CH}_3\text{CHO}]^{3/2}. Molecularity is the number of species colliding in an elementary step, it's theoretical, and it's always a whole number (1, 2 or 3, and 3 is rare). It has no meaning for the overall complex reaction.

  • [+]Pooja Iyer12thSample· 29d ago

    Mnemonic: Order = O for Observed (from the experiment). Molecularity = M for Mechanism step (collision count). When a reaction goes in several steps, the slowest step (the rate-determining step) decides the rate law. If the slow step is an elementary step, its molecularity gives its order.

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